Countable or Uncountable…That is the question!

advertisement

Countable or Uncountable…That
is the question!
REVIEW
• Countable
– Empty set, finite set or countably infinite
• Countably Infinite
– The set is a non-empty, non-finite set, and
there exists a bijection between N and the
set.
• Uncountable
– Not countable
HOMEWORK
Solutions
(1) Yes
The function f(n) = 2n is the desired bijection.
(2) Yes
The function f(n) =
desired bijection.
-n/2
if n is even
(n-1)/2 if n is odd
is the
Are the Rational Numbers Countable?
• What do we know about rational
numbers?
••••••-
VOTING
Are the Rational Numbers countable?
(A) YES
(B) NO
(C) UNSURE
What about the interval (0,1)
• What do we know about this interval?
••••••-
VOTING
Reals in the interval (0,1) countable?
(A) YES
(B) NO
(C) UNSURE
Let’s prove some things to attack
these questions!
• If A & B are disjoint countably infinite
sets then AυB is countable.
Proof
• Since A is countable there exists a bijection
f : N  A such that f (i) = ai
• Since B is countable there exists a bijection
g : N  B such that g (i) = bi
• Construct a function h (i) that orders the
elements of A and B in the following way:
a1, b1, a2, b2, a3, b3, . . .
Our function h (i)
• h (i) =
• Why is h a bijection?
RECORD these observations on your worksheet
Let’s prove some things to attack
these questions!
• If A & B are disjoint countably infinite
sets then AυB is countable.
• If A is a countably infinite set and B is a subset of A then
B is countable.
If A is a countably infinite set and B
is a subset of A then B is countable.
Case I: If B is the empty set or a finite set then B is
countable.
Case II: B is an infinite set
Since A is countable we can write the elements of A in the
order a1, a2, a3, . . .
If B is a subset of A then an infinite number of elements in
the above sequence are elements of B.
Thus the elements of B form a subsequence (c1, c2, c3,. . .)
of the sequence a1, a2, a3, . . ., thus we may order the
elements of B as b1, b2, b3, . . . where bk = ck and the
function f (i) = bi is a bijection between N and B
Let’s prove some things to attack
these questions!
• If A & B are disjoint countably infinite
sets then AυB is countable.
• If A is a countably infinite set and B is a subset of A then
B is countable.
• How does (N x N) relate to Q+ υ {0} in size?
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0,2
1,2
2,2
3,2
. . .
0,1
1,1
2,1
3,1
. . .
0,0
1,0
2,0
3,0
. . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
f(1) = (0,0)
.
f(2) = (1,0)
0,2
1,2
2,2
3,2
. . .
f(3) = (0,1)
f(4) = (0,2)
f(5) = (1,1)
0,1
1,1
2,1
3,1
. . .
f(6) = (2,0)
f(7) = (3,0)
0,0
1,0
2,0
3,0
. . .
f(8) = (2,1)
f(9) = (1,2)
.
.
.
Prove It!
• Now that we know that NxN is countable
we can show that Q is countable.
• Use the facts we have deduced to show
that Q is countable
Proof that Q is countable
• We know that Q+ υ {0} can be thought of
as a subset of NxN
• Similarly Q- can be thought of as a subset
of NxN
• Q+ υ {0} and Q- are countable because they
are subsets of a countable set.
• We have shown that the union of two
countable sets is also countable so
(Q+ υ {0}) υ Q- = Q is countable
Hey! Q is countable!
• Does this change your mind about the real
numbers in the interval (0,1) being
countable/uncountable?
(A) YES
(B) NO
(C) UNSURE
Cantor’s Diagonalization Argument
• Cantor proved that the interval of real
numbers (0,1) is… UNCOUNTABLE!!!
• We start by noting that each real number in the
interval (0,1) has a unique decimal
representation of the form 0.d1d2d3d4… (where
each di is a number from 0-9). And where
decimals with period 1 cannot repeat with the
number 9.
Proof (by contradiction)
• Assume that f is a bijection from N  (0,1). Then we may say:
f(1)
f(2)
f(3)
f(4)
=
=
=
=
A = 0.a1a2a3a4…
B = 0.b1b2b3b4…
(Where A,B,C,D are distinct real numbers in (0,1))
C = 0.c1c2c3c4…
D = 0.d1d2d3d4…
.
.
.
Choose a digit from 0 to 8 and we will call this a’1 such that a’1 ≠a1
Similarly choose a digit from 0 to 8 for b’1 such that b’2≠ b2
Similarly choose c’3≠ c3 and d’4≠ d4 . . .
Clearly f does not map a natural number to the real number
0.a’1b’2c’3d’4…
So f is not a bijection. Contradiction!
• This proof show us that the interval (0,1)
is actually a larger size of infinity than the
natural numbers.
HENCE (0,1) is a larger
size of infinity than Q as well!
Homework
• Question #1
– Today we identified two kinds of infinity: the size of the natural
numbers and the size of the interval (0,1).
– Show that there is a bijection between the interval (0,1) and the
set of real numbers.
– What does the existence of this bijection imply about these two
sets?
• Question #2
– Can you find a different size of infinity? That is a set that cannot
be put into a bijection with N or the interval (0,1).
– To help you with this problem research the findings of Paul
Cohen.